Can anyone solve the differential equation of x(1+y^2)^1/2dx=y(1+x^2)^1/2dy?

4 Answers

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  • rotchm
    Lv 7
    1 month ago

    Hint: From highschool algebra, do you see that this is separable? 

    Separate it first.  What do you get? Answer that & we will proceed. 

    Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. Too late!

  • 1 month ago

    x√( 1 + y² ) dx = y√( 1 + x² ) dy

    x / √( 1 + x² ) dx = y / √( 1 + y² ) dy

    ∫x / √( 1 + x² ) dx = ∫ y / √( 1 + y² ) dy

    LHS:

    let u = ( 1 + x² )  such that 

    du/dx = 2x

    dx = du/(2x)

    ∴ 

    ∫ x / √( 1 + x² ) dx 

    = ∫ ( x /√u )  *  du/(2x)

    = ½ ∫ u^(-½) du

    = √u

    = √( 1 + x² ) + constant 

    RHS:

    ∴ 

    ∫ y / √( 1 + y² ) dy 

    = √( 1 + y² ) + constant 

    THUS:

    √( 1 + y² ) = √( 1 + x² ) + C 

    1 + y²  = [ √( 1 + x² ) + C ]²

    y² =  [ √( 1 + x² ) + C ]² - 1

    y² = C² + x² + 2C√( x² + 1 )

    y = ±√[ C² + x² + 2C√( x² + 1 ) ]

  • 1 month ago

    The answer is as follows:

    Attachment image
  • Looks separable to me

    x * dx / (1 + x^2)^(1/2) = y * dy / (1 + y^2)^(1/2)

    u = (1 + x^2)^(1/2)

    u^2 = 1 + x^2

    2u * du = 2x * dx

    u * du = x * dx

    u * du / u =>

    du

    v = (1 + y^2)^(1/2)

    v^2 = 1 + y^2

    2v * dv = 2y * dy

    v * dv = y * dy

    y * dy / (1 + y^2)^(1/2)

    v * dv / v =>

    dv

    Now we have:

    du = dv

    Integrate

    u + C = v

    (1 + x^2)^(1/2) + C = (1 + y^2)^(1/2)

    You can expand from there if you want.

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